$\sum\limits_{k=1}^{2 }{{(3-k)}}=$
Explanation: What is the question asking for? $\sum$ tells us to find the sum. The question is asking for the sum of the values of $3-k$ from $k = 1$ to $k = 2$. Evaluating $\begin{aligned} \sum\limits_{k=1}^{2 }{({3-k})}&= (3-1) + (3-2) \\\\ &= 2 + 1 \\\\ &= 3\end{aligned}$ The answer $\sum\limits_{k=1}^{2 }{({3-k})}=3$